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Suppose m = 2 + 6i, and | m + n | = 3√10, where n is a complex number.

a) what is the minimum value of the modulus of n?

b) provide one example of the complex number, n.


For a, I got √10 for the minimum value of modulus n.

Here's my work:


| 2 + 6i + n | = 3√10
- 2 + 6i - 2 + 6i
n = 3√10 - 2 + 6i

I then found the modulus of -2+6 and got 2√10, and replaced it, so I can solve the rest of the equation.

n = 3√10 - 2√10
n = √10


For b, I'm not really sure what it means by providing an example. I tend to overthink things, so if someone could maybe explain and help me, that would be appreciated! (And if I somehow got a wrong, please tell me what I did wrong! Thank :)

Respuesta :

facundo3141592

Answer: a) √50

b) n = 1 + 7i

Step-by-step explanation:

first, the modulus of a complex number z = a + bi is

IzI = √(a^2 + b^2)  

The fact that n is complex does not mean that n doesn't has a real part, so we must write our numbers as:

m = 2 + 6i

n = a  + bi

Im + nI = 3√10

Im + n I = √(a^2 + b^2 + 2^2 + 6^2)= 3√10

            = √(a^2 + b^2 + 40) = 3√10

             a^2 + b^2 + 40 = 3^2*10 = 9*10 = 90

             a^2 + b^2 = 90 - 40 = 50

            √(a^2 + b^2 ) = InI = √50

The modulus of n must be equal to the square root of 50.

now we can find any values a and b such a^2 + b^2 = 50.

for example, a = 1 and b = 7

1^2 + 7^2 = 1 + 49  = 50

Then a possible value for n is:

n = 1 + 7i

msm555

Answer:

the modulus of a complex number z = a + bi is:

Izl= √(a²+b²)

The fact that n is complex does not mean that n doesn't has a real part, so we must write our numbers as:

m = 2 + 6i

n = a + bi

Im + nl = 3√10

√(a² + b²+ 2²+ 6²)= 3√10

√(a^2 + b^2 + 40) = 3√10

squaring both side

a²+b²+40 = 3^2*10 = 9*10 =90

a²+b²= 90 - 40

a²+b²=50

So,

|n|=√(a^2 + b^2) = √50

The modulus of n must be equal to the square root of 50.

now

values a and b such

a^2 + b^2 = 50.

for example, a = 5 and b = 5

5²+5²=25+25= 50

Then a possible value for n is:

n = 5+5i

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